Vertex relabeling preserves the uniform regular graph law

By F3, each vertex has a unique preimage under $\sigma$; mapping a vertex to that preimage defines the inverse permutation ${\sigma }^{-1}$. Relabeling preserves distinct endpoints, and $v\mapsto \sigma v$ bijects the neighbors of $u$ with the neighbors of $\sigma u$. Thus $\sigma G$ remains simple and $d$-regular. Relabeling by ${\sigma }^{-1}$ inverts this construction, so $T_{\sigma }:G\mapsto \sigma G$ is a bijection of the finite set ${\mathrm{\Omega }}_{n,d}$. For every event $A\subseteq {\mathrm{\Omega }}_{n,d}$, its restriction is a bijection from $A$ onto $T_{\sigma }(A)$, so $\mathbb{P}(T_{\sigma }(A))=\mathrm{\mid }{T}_{\sigma }(A)\mathrm{\mid }\mathrm{/}\mathrm{\mid }{\mathrm{\Omega }}_{n,d}\mathrm{\mid }=\mathrm{\mid }A\mathrm{\mid }\mathrm{/}\mathrm{\mid }{\mathrm{\Omega }}_{n,d}\mathrm{\mid }=\mathbb{P}(A)$. Hence the uniform law is preserved.

If $n\ge 3$, applying $\sigma$ to a vertex listing defining a Hamiltonian cycle gives a listing defining a Hamiltonian cycle of $\sigma G$. Applying ${\sigma }^{-1}$ proves the converse. If $n=1,2$, no simple cycle exists by F5, so neither graph is Hamiltonian under the stated convention. Thus Hamiltonicity is preserved for every allowed $n$. A relabeling-invariant property is preserved by its defining invariance.

If candidate cycles $C,D$ exist, then $n\ge 3$. Choose listings $(v_1,\dots ,v_n)$ and $(w_1,\dots ,w_n)$ defining them. Set $\sigma (v_i)=w_i$ for every $i$. Each listing contains every vertex exactly once, so this defines a bijection of $[n]$, with inverse $w_i\mapsto v_i$, hence a permutation. It carries every consecutive edge, including the closing edge, to the corresponding edge of $D$, giving $\sigma C=D$. For $n=1,2$ the assertion about pairs of candidate cycles is vacuous.